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Ozone, \(\mathrm{O}_{3}\), decomposes to molecular oxygen in the stratosphere according to the reaction \(2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g) .\) Would an increase in pressure favor the formation of ozone or of oxygen?

Short Answer

Expert verified
An increase in pressure would favor the formation of ozone, \(\mathrm{O}_{3}\), due to the reaction seeking to counteract the change by shifting towards the side with fewer moles of gas, according to Le Chatelier's principle.

Step by step solution

01

Examine the balanced chemical equation

First, let us examine the balanced chemical equation for the decomposition reaction of ozone: \(2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g) \) This equation tells us that 2 moles of ozone, \(\mathrm{O}_{3}\), decompose into 3 moles of molecular oxygen, \(\mathrm{O}_{2}\).
02

Analyze the reaction with respect to pressure

Now, we'll consider the effect of pressure on the reaction. If the pressure is increased, the system will react by trying to decrease the pressure, according to Le Chatelier's principle. This can be achieved by favoring the side of the reaction with fewer moles of gas. In our case, there are more moles of gas on the right side (3 moles of \(\mathrm{O}_{2}\)) than on the left side (2 moles of \(\mathrm{O}_{3}\)). Therefore, the system will favor a reaction that goes in the direction of fewer moles of gases to counteract the increased pressure.
03

Determine the favored side

Given that the system will favor the side of the reaction with fewer moles of gas when the pressure is increased, we can determine that the formation of ozone, \(\mathrm{O}_{3}\), will be favored under increased pressure. This indicates that increasing the pressure will shift the equilibrium towards the left side of the reaction, favoring formation of ozone rather than molecular oxygen. Hence, an increase in pressure would favor the formation of ozone.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ozone decomposition
Ozone (\( \mathrm{O}_{3} \)) is a molecule composed of three oxygen atoms. It plays a crucial role in absorbing harmful UV radiation in the stratosphere. The breakdown or decomposition of ozone into molecular oxygen (\( \mathrm{O}_{2} \)) is a vital chemical reaction for maintaining the balance of ozone in the atmosphere.

In the given reaction — \( 2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g) \) — two moles of ozone break down to form three moles of oxygen gas.
This decomposition is part of the ozone-oxygen cycle which protects life on earth by filtering solar radiation. However, this balance can be affected by the introduction of chemicals or changes in environmental conditions, such as pressure or temperature. Understanding this equilibrium is essential for predicting how changes in the environment might impact ozone levels.
Pressure effects on equilibrium
Pressure plays a significant role in determining the direction of chemical reactions, especially those involving gases. In a gaseous system, an increase in pressure usually shifts the equilibrium towards the side of the reaction with fewer gas molecules, according to Le Chatelier's Principle. This principle states that a system at equilibrium will adjust to counteract changes and restore balance.

In the context of ozone decomposition, as shown in the reaction \( 2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g) \), an increase in pressure will shift the equilibrium to favor the production of more ozone (\( \mathrm{O}_{3} \)), since it consists of fewer total moles of gas compared to the product side.
  • Left side: 2 moles of gas (\( \mathrm{O}_{3} \))
  • Right side: 3 moles of gas (\( \mathrm{O}_{2} \))
Increasing the pressure will therefore encourage the reaction to shift to the left side, reducing the number of gas molecules and decreasing the overall pressure, thus, leading to more ozone being formed.
Mole concept
The mole concept is a fundamental concept in chemistry which is used to express amounts of a chemical substance. One mole corresponds to \( 6.022 \times 10^{23} \) entities, which is the Avogadro's number.

In the ozone decomposition, understanding moles helps in balancing chemical reactions and determining the quantities of reactants and products.
For example, the chemical equation\( 2 \mathrm{O}_{3} \rightarrow 3 \mathrm{O}_{2} \)demonstrates a conversion of two moles of ozone (each mole consisting of \( 6.022 \times 10^{23} \) molecules of ozone) into three moles of oxygen gas.
  • If you start with 2 moles of ozone, you’ll end up with 3 moles of oxygen.
  • This type of stoichiometry calculation uses the mole ratio from the coefficients of the balanced equation.
Using the mole concept facilitates precise measurement and prediction in chemical reactions, enabling chemists to predict how much product results from given quantities of reactants.

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Most popular questions from this chapter

Ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with halogens \(\left(\mathrm{X}_{2}\right)\) by the following reaction: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g) $$ The following figures represent the concentrations at equilibrium at the same temperature when \(\mathrm{X}_{2}\) is \(\mathrm{Cl}_{2}\) (green), \(\mathrm{Br}_{2}\) (brown), and \(\mathrm{I}_{2}\) (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3\(]\)

Consider the following equilibrium, for which \(K_{p}=0.0752\) at $$\begin{array}{l} 480{ }^{\circ} \mathrm{C}: \\ \quad 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \end{array}$$ (a) What is the value of \(K_{p}\) for the reaction \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathrm{c})\) What is the value of \(K_{c}\) for the reac- tion in part (b)?

At \(900 \mathrm{~K}\) the following reaction has \(K_{p}=0.345\) : $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ In an equilibrium mixture the partial pressures of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

When the following reactions come to equilibrium, does he equilibrium mixture contain mostly reactants or mostly broducts? a) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) ; K_{c}=1.5 \times 10^{-10}\) b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) ; K_{p}=2.5 \times 10^{9}\)

At \(1000 \mathrm{~K}, K_{p}=1.85\) for the reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathbf{b})\) What is the value of \(K_{p}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) ? (c) What is the value of \(K_{c}\) for the reaction in part (b)?

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